Q.1 Solve

(a)\(\frac{2}{3} + \frac{1}{7} \)

(b)\(\frac{3}{10} + \frac{7}{15} \)

(c) \(\frac{4}{9} + \frac{2}{7} \)

(d) \(\frac{5}{7} + \frac{1}{3} \)

(e) \(\frac{2}{5} + \frac{1}{6} \)

(f) \(\frac{4}{5} + \frac{2}{3} \)

(g) \(\frac{3}{4} - \frac{1}{3} \)

(h) \(\frac{5}{6} - \frac{1}{3} \)

(i) \(\frac{2}{3} + \frac{3}{4} + \frac{1}{2} \)

(j) \(\frac{1}{2} + \frac{1}{3} + \frac{1}{6} \)

(k) \(1\frac{1}{3} + 3\frac{2}{3} \)

(l) \(4\frac{2}{3} + 3\frac{1}{4} \)

(m) \(\frac{16}{5} - \frac{7}{5} \)

(n) \(\frac{4}{3} - \frac{1}{2} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a)\(\frac{2}{3} + \frac{1}{7} \)

LCM of 7 and 3 = 21

\(\frac{2×7}{3×7} + \frac{1×3}{7×3} \)

= \(\frac{14}{21} + \frac{3}{21} \)

= \(\frac{17}{21} \)

(b)\(\frac{3}{10} + \frac{7}{15} \)

LCM of 10 and 15 = 30

\(\frac{3×3}{10×3} + \frac{7×2}{15×2} \)

= \(\frac{9}{30} + \frac{14}{30} \)

= \(\frac{23}{30} \)

(c) \(\frac{4}{9} + \frac{2}{7} \)

LCM of 9 and 7 = 63

\(\frac{4×7}{9×7} + \frac{2×9}{7×9} \)

= \(\frac{28}{63} + \frac{18}{63} \)

= \(\frac{46}{63} \)

(d) \(\frac{5}{7} + \frac{1}{3} \)

LCM of 7 and 3 = 21

\(\frac{5×3}{7×3} + \frac{1×7}{3×7} \)

= \(\frac{15}{21} + \frac{7}{21} \)

= \(\frac{22}{21} \)

(e) \(\frac{2}{5} + \frac{1}{6} \)

LCM of 5 and 6 = 30

\(\frac{2×6}{5×6} + \frac{1×5}{6×5} \)

= \(\frac{12}{30} + \frac{5}{30} \)

= \(\frac{17}{30} \)

(f) \(\frac{4}{5} + \frac{2}{3} \)

LCM of 5 and 3 = 15

\(\frac{4×3}{5×3} + \frac{2×5}{3×5} \)

= \(\frac{12}{15} + \frac{10}{15} \)

= \(\frac{22}{15} \)

(g) \(\frac{3}{4} - \frac{1}{3} \)

LCM of 4 and 3 = 12

\(\frac{3×3}{4×3} - \frac{1×4}{3×4} \)

= \(\frac{9}{12} - \frac{4}{12} \)

= \(\frac{5}{12} \)

(h) \(\frac{5}{6} - \frac{1}{3} \)

LCM of 6 and 3 = 6

\(\frac{5}{6} - \frac{1×2}{3×2} \)

= \(\frac{5}{6} - \frac{2}{6} \)

= \(\frac{3}{6} \) = \(\frac{1}{2} \)

(i) \(\frac{2}{3} + \frac{3}{4} + \frac{1}{2} \)

LCM of 3, 4 and 2 = 12

\(\frac{2×4}{3×4} + \frac{3×3}{4×3} + \frac{1×6}{2×6} \)

\(\frac{8}{12} + \frac{9}{12} + \frac{6}{12} \)

\(\frac{23}{12} \)

(j) \(\frac{1}{2} + \frac{1}{3} + \frac{1}{6} \)

LCM of 2, 3 and 6 = 6

\(\frac{1×3}{2×3} + \frac{1×2}{3×2} + \frac{1}{6} \)

\(\frac{3}{6} + \frac{2}{6} + \frac{1}{6} \)

\(\frac{6}{6} \) = 1

(k) \(1\frac{1}{3} + 3\frac{2}{3} \)

= \(\frac{4}{3} + \frac{11}{3} \)

= \(\frac{15}{3} \) = 5

(l) \(4\frac{2}{3} + 3\frac{1}{4} \)

= \(\frac{14}{3} + \frac{13}{4} \)

LCM of 3 and 4 = 12

\(\frac{14×4}{3×4} + \frac{13×3}{4×3} \)

= \(\frac{56}{12} + \frac{39}{12} \)

= \(\frac{95}{12} \) \)

(m) \(\frac{16}{5} - \frac{7}{5} \)

\(\frac{9}{5} \) = \(1\frac{4}{5} \)

(n) \(\frac{4}{3} - \frac{1}{2} \)

LCM of 3 and 2 = 6

\(\frac{4×2}{3×2} - \frac{1×3}{2×3} \)

= \(\frac{8}{6} - \frac{3}{6} \)

= \(\frac{5}{6} \) \)

Q.2 Sarita bought \(\frac{2}{5} \) metre of ribbon and Lalita \(\frac{3}{4} \) metre of ribbon. What is the total length of the ribbon they bought?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Length of ribbon bought by Sarita = \(\frac{2}{5} \) metre

Length of ribbon bought by Lalita = \(\frac{3}{4} \) metre

\(\therefore \) Length of ribbon bought by Sarita and Lalita
\(\frac{2}{5} + \frac{3}{4} \)

LCM of 5 and 4 = 20

\(\frac{2×4}{5×4} + \frac{3×5}{4×5} \)

= \(\frac{8}{20} + \frac{15}{20} \)

= \(\frac{23}{20} \) = \(1\frac{3}{20} \) meter

Q.3 Naina was given \(1\frac{1}{2} \) piece of cake and Najma was given \(1\frac{1}{3} \) piece of cake. Find the total amount of cake was given to both of them.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Fraction of cake Naina got = \(1\frac{1}{2} \) = \(\frac{3}{2} \)

Fraction of cake Najma got = \(1\frac{1}{3} \)= \(\frac{4}{3} \)

Total amount of cake given to both of them =

\(\frac{3}{2} + \frac{4}{3} \)

LCM of 2 and 3 = 6

\(\frac{3×3}{2×3} + \frac{4×2}{3×2} \)

= \(\frac{9}{6} + \frac{8}{6} \)

= \(\frac{17}{6} \) = \(2\frac{5}{6} \)

Q. 4. Fill in the boxes:

(a) \(\square - \frac{5}{8} = \frac{1}{4} \)

(b) \(\square - \frac{1}{5} = \frac{1}{2} \)

(c) \(\frac{1}{2} - \square = \frac{1}{6} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\square - \frac{5}{8} = \frac{1}{4} \)

Or, \(\square = \frac{1}{4} + \frac{5}{8} \)

= \( \frac{1×2}{4×2} + \frac{5}{8} \)

= \( \frac{2}{8} + \frac{5}{8} \) = \(\frac{7}{8}

(b) \(\square - \frac{1}{5} = \frac{1}{2} \)

Or, \(\square = \frac{1}{2} + \frac{1}{5} \)

= \( \frac{1×5}{2×5} + \frac{1×2}{5×2} \)

= \( \frac{5}{10} + \frac{2}{10} \) = \(\frac{7}{10} \)

(C) \(\frac{1}{2} - \square = \frac{1}{6} \)

Or, \(\frac{1}{2} - \frac{1}{6} = \square \)

Or, \(\square = \frac{1×3}{2×3} - \frac{1}{6} \)

Or, \(\square = \frac{3}{6} - \frac{1}{6} \)

Or, \(\square = \frac{2}{6} \) = \(\frac{1}{3} \)

Q6 A piece of wire \(\frac{7}{8} \) metre long broke into two pieces. One piece was \(\frac{1}{4} \) metre long. How long is the other piece?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Total length of wire = \(\frac{7}{8} \) metre

Length of one piece of wire = \(\frac{1}{4} \)metre

Length of other piece of wire = Length of the original wire and this one piece of wire

= \(\frac{7}{8} - \frac{1}{4} \)

= \(\frac{7}{8} - \frac{1×2}{4×2} \)

= \(\frac{7}{8} - \frac{2}{8} \)

= \(\frac{5}{8} \)

Q.7 Nandini’s house is \(\frac{9}{10} \) km from her school. She walked some distance and then took a bus for \(\frac{1}{2} \) km to reach the school. How far did she walk?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Distance of the school from house = \(\frac{9}{10} \) km

Distance she travelled by bus = \(\frac{1}{2} \) km

Distance walked by Nandini = Total distance of the school – Distance she travelled by bus

= \(\frac{9}{10} - \frac{1}{2} \)

= \(\frac{9}{10} - \frac{1×5}{2×5} \)

= \(\frac{9}{10} - \frac{5}{10} \)

= \(\frac{4}{10} \) = \(\frac{2}{5} \) km

Q.8 Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is \(\frac{5}{6} \) th full and Samuel’s shelf is \(\frac{2}{5} \) th full. Whose bookshelf is more full? By what fraction?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Fraction of Asha’s bookshelf = \(\frac{5}{6} \)

Fraction of Samuel’s bookshelf = \(\frac{2}{5} \)

\(\frac{5}{6} \) and \(\frac{2}{5} \)

By cross multiplication

5×5 and 6×2

25 > 12

\(\frac{5}{6} \) > \(\frac{2}{5} \)

\(\therefore \) Asha’s bookshelf is more full than Samuel’s bookshelf

Difference = \(\frac{5}{6} \) - \(\frac{2}{5} \)

= \(\frac{5×5}{6×5} \) - \(\frac{2×6}{5×6} \)

= \(\frac{25}{30} \) - \(\frac{12}{30} \)

= \(\frac{13}{30} \)

Q.9 Jaidev takes\(2\frac{1}{5} \) minutes to walk across the school ground. Rahul takes\(\frac{7}{4} \) minutes to do the same. Who takes less time and by what fraction?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Time taken by Jaidev to walk across the school ground =\(2\frac{1}{5} \) = \(\frac{11}{5} \) minutes

Time taken by Rahul to walk across the school ground =
\(\frac{7}{4} \) minutes

\(\frac{11}{5} \) and \(\frac{7}{4} \)

By cross multiplication

11×4 and 5×7

44 > 35

\(\frac{11}{5} \) > \(\frac{7}{4} \)

\(\therefore \) Rahul takes less time than Jaidev to walk across the school ground

Difference = \(\frac{11}{5} \) - \(\frac{7}{4} \)

= \(\frac{11×4}{5×4} \) - \(\frac{7×5}{4×5} \)

= \(\frac{44}{20} \) - \(\frac{35}{20} \)

= \(\frac{9}{20} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(i) Number of parts = 4

Shaded portion = 2

\(\therefore \) Fraction = \(\frac{2}{4} \)

(ii) Number of parts = 9

Shaded portion = 8

\(\therefore \) Fraction = \(\frac{8}{9} \)

(iii) Number of parts = 8

Shaded portion = 4

\(\therefore \) Fraction = \(\frac{4}{8} \)

(iv) Number of parts = 4

Shaded portion = 1

\(\therefore \) Fraction = \(\frac{1}{4} \)

(v) Number of parts = 7

Shaded portion = 3

\(\therefore \) Fraction = \(\frac{3}{7} \)

(vi) Number of parts = 12

Shaded portion = 3

\(\therefore \) Fraction = \(\frac{3}{12} \)

(vii) Number of parts = 10

Shaded portion = 10

\(\therefore \) Fraction = \(\frac{10}{10} \)

(viii) Number of parts = 9

Shaded portion = 4

\(\therefore \) Fraction = \(\frac{4}{9} \)

(ix) Number of parts = 8

Shaded portion = 4

\(\therefore \) Fraction = \(\frac{4}{8} \)

(x) Number of parts = 2

Shaded portion = 1

\(\therefore \) Fraction = \(\frac{1}{2} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(i) The parts are not equal

Hence, this is not \(\frac{1}{2} \)

(ii) the parts are not equal

Hence, Shaded portion is not \(\frac{1}{4} \)

(iii) the parts are not equal

Hence, Shaded portion is not \(\frac{3}{4} \)

Q.6 Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.

(a) How can Arya divide his sandwiches so that each person has an equal share?

(b) What part of a sandwich will each boy receive?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) Arya has divided the sandwich into 3 equal parts. So each person will get one part.

(b) Each boy receive \(\frac{1}{3} \)

Q.7 Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Total number of dresses to be dyed = 30

Number of dresses finished = 20

\(\therefore \) Required fraction = \(\frac {20}{30} = \frac{2}{3} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Natural numbers between 2 and 12 are;

2,3,4, 5, 6, 7, 8, 9, 10,11, 12

No. of given natural numbers = 11

No. of prime numbers = 5

Hence, fraction = \(\frac{5}{11} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Natural numbers from 102 to 113 are;

102,103,104,105,106, 107,108, 109,110, 111, 112,113

Total number of given natural numbers = 12

Prime numbers are 103, 107, 109, 113

Hence, fraction = \(\frac{4}{12} \) = \(\frac{1}{3} \)

Q.11 Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Number of CDs bought by her from the market = 3

Number of CDs received as gifts = 5

Therefore Total number of CDs = 3 + 5 = 8

Therefore Fraction of CDs (bought) = \(\frac{3}{8} \)

and the fraction of CDs (gifted) = \(\frac{5}{8} \)

Q.1 Draw number lines and locate the points on them:

(a)\(\frac{1}{2} , \frac{1}{ 4} , \frac{3}{4} , \frac{4}{4} \)

(b) \(\frac{1}{8} , \frac{2}{ 8} , \frac{3}{8} , \frac{7}{8} \)

(c) \(\frac{2}{5} , \frac{3}{ 5} , \frac{8}{5} , \frac{4}{5} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Q.2 Express the following as mixed fractions:

(a)\(\frac{20}{3} \)

(b) \(\frac{11}{5} \)

(c) \(\frac{17}{7} \)

(d) \(\frac{28}{5} \)

(e) \(\frac{19}{6} \)

(f) \(\frac{35}{9} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a)\(\frac{20}{3} \) = \( 6 \frac{2}{3} \)

(b) \(\frac{11}{5} \) = \( 2 \frac{1}{5} \)

(c) \(\frac{17}{7} \) = \( 2 \frac{3}{7} \)

(d) \(\frac{28}{5} \) = \( 5 \frac{3}{5} \)

(e) \(\frac{19}{6} \) = \( 3 \frac{1}{6} \)

(f) \(\frac{35}{9} \) = \( 3 \frac{8}{9} \)

Q.3 Express the following as improper fractions:

(a) \(7\frac{3}{4} \)

(b) \(5\frac{6}{7} \)

(c) \(2\frac{5}{6} \)

(d) \(10\frac{3}{5} \)

(e) \(9\frac{3}{7} \)

(f) \(8\frac{4}{9} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{(7 × 4 + 3)}{4} = \frac{31}{4} \)

\(\therefore \) The improper form is \(\frac{31}{4} \)

(b) \(\frac{(5 × 7 + 6)}{7} = \frac{41}{7} \)

\(\therefore \) The improper form is \(\frac{41}{7} \)

(c) \(\frac{(2 × 6 + 5)}{6} = \frac{17}{6} \)

\(\therefore \) The improper form is \(\frac{17}{6} \)

(d) \(\frac{(10 × 5 + 3)}{5} = \frac{53}{5} \)

\(\therefore \) The improper form is \(\frac{53}{5} \)

(e) \(\frac{(9 × 7 + 3)}{7} = \frac{66}{7} \)

\(\therefore \) The improper form is \(\frac{66}{7} \)

(f) \(\frac{(8 × 9 + 4)}{9} = \frac{76}{9} \)

\(\therefore \) The improper form is \(\frac{76}{9} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a)

(i) The shaded portion is \(\frac{1}{2} \)

(ii) The shaded portion is\(\frac{2}{4} = \frac{1}{2} \)

(iii) The shaded portion is \(\frac{3}{6} = \frac{1}{2} \)

(iv) The shaded portion is \(\frac{4}{8} = \frac{1}{2} \)

Hence, all fractions are equivalent.

(b)

(i) The shaded portion is \(\frac{4}{12} = \frac{1}{3} \)

(ii) The shaded portion is\(\frac{3}{9} = \frac{1}{3} \)

(iii) The shaded portion is \(\frac{2}{6} = \frac{1}{3} \)

(iv) The shaded portion is \(\frac{6}{15} = \frac{2}{5} \)

Hence, all fractions are not equivalent.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a)\(\frac{1}{2} \)

(b) \(\frac{4}{6} \) = \(\frac{2}{3} \)

(c) \(\frac{3}{9} \) = \(\frac{1}{3} \)

(d) \(\frac{2}{8} \) = \(\frac{1}{4} \)

(e) \(\frac{3}{4} \)

(i) \(\frac{6}{18} \) = \(\frac{1}{3} \)

(ii) \(\frac{4}{8} \) = \(\frac{1}{2} \)

(iii) \(\frac{12}{16} \) = \(\frac{3}{4} \)

(iv) \(\frac{8}{12} \) = \(\frac{2}{3} \)

(v) \(\frac{4}{16} \) = \(\frac{1}{4} \)

The following are the equivalent fractions

(a) and (ii)

(b) and (iv)

(c) and (i)

(d) and (v)

(e) and (iii)

Q.3 Replace \(\square \)in each of the following by the correct number:

(a) \(\frac{2}{7} = \frac{8}{\square } \)

(b) \(\frac{5}{8} = \frac{10}{\square } \)

(c) \(\frac{3}{5}= \frac{\square }{20} \)

(d) \(\frac{45}{60} = \frac{15}{\square } \)

(e) \(\frac{18}{24}= \frac{\square }{4} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{2}{7} = \frac{8}{\square } \)

\(\rightarrow \) \(\frac{2×4}{7×4} = \frac{8}{28} \)

(b) \(\frac{5}{8} = \frac{10}{\square } \)

\(\rightarrow \) \(\frac{5×2}{8×2} = \frac{10}{16} \)

(c) \(\frac{3}{5}= \frac{\square }{20} \)

\(\rightarrow \) \(\frac{3×4}{5×4} = \frac{12}{20} \)

(d) \(\frac{45}{60} = \frac{15}{\square } \)

\(\rightarrow \) \(\frac{45÷3}{60÷3} = \frac{15}{20} \)

(e) \(\frac{18}{24}= \frac{\square }{4} \)

\(\rightarrow \) \(\frac{18÷6}{24÷6} = \frac{3}{4} \)

Q.4 Find the equivalent fraction of \(\frac{3}{5} \)having

(a) denominator 20

(b) numerator 9

(c) denominator 30

(d) numerator 27

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) We require denominator 20

Let M be the numerator of the fractions

\(\therefore \frac{M}{20} = \(\frac{3}{5} \)

5 × M = 20 × 3

M = (20 × 3) ÷ 5 = 12

Therefore the required fraction is \(\frac{12}{20} \)

(b) We require numerator 9

Let N be the denominator of the fractions

\(\therefore \frac{9}{N} = \(\frac{3}{5} \)

3 × N = 9 × 5

N = (9 × 5) ÷ 3 = 15

Therefore the required fraction is \(\frac{9}{15} \)

(c) We require denominator 30

Let D be the numerator of the fraction

\(\therefore \frac{D}{30} = \(\frac{3}{5} \)

5 × D = 3 × 30

D = (3 × 30) ÷ 5

= 18

Therefore the required fraction is \(\frac{18}{30} \)

(d) We require numerator 27

Let X be the denominator of the fraction

\(\therefore \frac{27}{X} = \(\frac{3}{5} \)

3 × X = 5 × 27

X = (5 × 27) ÷ 3 = 45

Therefore the required fraction is \(\frac{27}{45} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) Given numerator = 9

\(\therefore \frac{9}{X} = \frac{36}{48} \)

X × 36 = 9 × 48

X = (9 × 48) / 36

X = 12

Hence, the equivalent fraction is \(\frac{9}{12} \)

(b) Given, denominator = 4

\(\therefore \frac{X}{4} = \frac{36}{48} \)

X × 48 = 4 × 36

X = (4 × 36) / 48 = 3

Hence, the equivalent fraction is \(\frac{3}{4} \)

Q.6 Check whether the given fractions are equivalent:

(a) \(\frac{5}{9} , \frac{30}{54} \)

(b) \(\frac{3}{10} , \frac{12}{50} \)

(c) \(\frac{7}{13} , \frac{5}{11} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) Given \(\frac{5}{9} \) and \(\frac{30}{54} \)

We have 5× 54 = 270

9 × 30 = 270

5 × 54 = 9 × 30

Hence, \(\frac{5}{9} \) and \(\frac{30}{54} \) are equivalent fractions

(b) Given \(\frac{3}{10} \) and \(\frac{12}{50} \)

We have 3 × 50 = 150

10 × 12 = 120

3 × 50 \(\neq \)10 × 12

Hence, \(\frac{3}{10} \) and \(\frac{12}{50} \) are not equivalent fractions

(c) Given \(\frac{7}{13} \) and \(\frac{5}{11} \)

We have 7 × 11 = 77

5 × 13 = 65

7 × 11 \(\neq \) 5 × 13

Hence, \(\frac{7}{13} \) and \(\frac{5}{11} \) are not equivalent fractions

Q.7 Reduce the following fractions to simplest form:

(a) \(\frac{48}{60} \)

(b) \(\frac{150}{60} \)

(c) \(\frac{84}{98} \)

(d) \(\frac{12}{52} \)

(e) \(\frac{7}{28} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{48}{60} \) = \(\frac{48÷12}{60÷12} \)

= \(\frac{4}{5} \)

(b) \(\frac{150}{60} \) = \(\frac{150÷30}{60÷30} \)

= \(\frac{5}{2} \)

(c) \(\frac{84}{98} \)= \(\frac{84÷14}{98÷14} \)

= \(\frac{6}{7} \)

(d) \(\frac{12}{52} \) = \(\frac{12÷4}{52÷4} \)

= \(\frac{3}{13} \)

(e) \(\frac{7}{28} \) = \(\frac{7÷7 }{28÷7 } \)

= \(\frac{1}{4} \)

Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Total number of pencils Ramesh had = 20

Number of pencils used by Ramesh = 10

\(\therefore \) Fraction = \(\frac{10}{20} = \frac{1}{2} \)

Total number of pencils Sheelu had = 50

Number of pencils used by Sheelu = 25

\(\therefore \) Fraction = \(\frac{25}{50} = \frac{1}{2} \)

Total number of pencils Jamaal had = 80

Number of pencils used by Jamaal = 40

\(\therefore \) Fraction = \(\frac{40}{80} = \frac{1}{2} \)

Yes, each has used up an equal fraction of pencils i.e \(\frac{1}{2} \)

Q.9 Match the equivalent fractions and write two more for each.

(i) \(\frac{250}{400} \) (a) \(\frac{2}{3} \)

(ii) \(\frac{180}{200} \) (b) \(\frac{2}{5} \)

(iii) \(\frac{660}{990} \) (c) \(\frac{1}{2} \)

(iv) \(\frac{180}{360} \) (d) \(\frac{5}{8} \)

(v) \(\frac{220}{550} \) (e) \(\frac{9}{10} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(i) \(\frac{250}{400} \) = (d) \(\frac{5}{8} \)

(ii) \(\frac{180}{200} \) = (e) \(\frac{9}{10} \)

(iii) \(\frac{660}{990} \) = (a) \(\frac{2}{3} \)

(iv) \(\frac{180}{360} \) = (c) \(\frac{1}{2} \)

(v) \(\frac{220}{550} \) = (b) \(\frac{2}{5} \)

Q.1 Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘’ between the fractions:

(c) Show\(\frac{2}{6} , \frac{4}{6} , \frac{8}{6} and \frac{6}{6} \) on the number line. Put appropriate signs between the fractions given.

\(\frac{5}{6} \square \frac{2}{6} , \frac{3}{6} \square 0, \frac{1}{6} \square \frac{6}{6} , \frac{8}{6} \square \frac{5}{6}

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{3}{8} , \frac{6}{8} , \frac{4}{8} and \frac{1}{8} \)

Ascending order = \(\frac{1}{8} < \frac{3}{8} < \frac{4}{8} < \frac{6}{8} \)

Decending order = \(\frac{6}{8} > \frac{4}{8} > \frac{3}{8} > \frac{1}{8} \)

(b) \(\frac{8}{9} , \frac{4}{9} , \frac{3}{9} and \frac{6}{9} \)

Ascending order = \(\frac{3}{9} < \frac{4}{9} < \frac{6}{9} < \frac{8}{9} \)

Decending order = \(\frac{8}{9} > \frac{6}{9} > \frac{4}{9} > \frac{3}{9} \)

(c)

\(\frac{5}{6} > \frac{2}{6} , \frac{3}{6} > 0, \frac{1}{6} < \frac{6}{6} , \frac{8}{6} > \frac{5}{6}

Q.2 Compare the fractions and put an appropriate sign.

(a)\( \frac{3}{6} \square \frac{5}{6} \)

(b) \( \frac{1}{7} \square \frac{1}{4} \)

(c) \( \frac{4}{5} \square \frac{5}{5} \)

(d)\( \frac{3}{5} \square \frac{3}{7} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a)\( \frac{3}{6} < \frac{5}{6} \)

If denominators are same than whose numerator is greater that fraction also greater.

(b) \( \frac{1}{7} \ < \frac{1}{4} \)

If denominators are same than whose numerator is greater that fraction also greater.

(c) \( \frac{4}{5} < \frac{5}{5} \)

If denominators are same than whose numerator is greater that fraction also greater.

(d)\( \frac{3}{5} > \frac{3}{7} \)

If denominators are same than whose numerator is greater that fraction also greater.

Q.4 Look at the figures and write ’<’, or ’>’ ’=’ between the given pairs of fractions.

(a) \(\frac{1}{6} \square \frac{1}{3} \)

(b) \(\frac{3}{4} \square \frac{2}{6} \)

(c) \(\frac{2}{3} \square \frac{2}{4} \)

(d) \(\frac{6}{6} \square \frac{3}{3} \)

(e) \(\frac{5}{6} \square \frac{5}{5} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{1}{6} < \frac{1}{3} \)

(b) \(\frac{3}{4} > \frac{2}{6} \)

(c) \(\frac{2}{3} > \frac{2}{4} \)

(d) \(\frac{6}{6} = \frac{3}{3} \)

(e) \(\frac{5}{6} < \frac{5}{5} \)

Q.5 How quickly can you do this? Fill appropriate sign. ( ‘<’, ‘=’, ‘>’)

(a) \(\frac{1}{2} \square \frac{1}{5} \)

(b) \(\frac{2}{4} \square \frac{3}{6} \)

(c) \(\frac{3}{5} \square \frac{2}{3} \)

(d) \(\frac{3}{4} \square \frac{2}{8} \)

(e) \(\frac{3}{5} \square \frac{5}{5} \)

(f) \(\frac{7}{9} \square \frac{3}{9} \)

(g) \(\frac{1}{4} \square \frac{2}{8} \)

(h) \(\frac{6}{10} \square \frac{4}{5} \)

(i) \(\frac{3}{4} \square \frac{7}{8} \)

(j) \(\frac{6}{10} \square \frac{3}{5} \)

(k) \(\frac{5}{7} \square \frac{15}{21} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{1}{2} \square \frac{1}{5} \)

\(\frac{1}{2} and \frac{1}{5} \)

5 × 1 \(\square \) 2 × 1

5 > 2

Hence, \(\frac{1}{2} > \frac{1}{5} \)

(b) \(\frac{2}{4} \square \frac{3}{6} \)

\(\frac{2}{4} and \frac{3}{6} \)

6 × 2 \(\square \) 4 × 3

12 = 12

Hence, \(\frac{2}{4} = \frac{3}{6} \)

(c) \(\frac{3}{5} \square \frac{2}{3} \)

\(\frac{3}{5} and \frac{2}{3} \)

3 × 3 \(\square \) 5 × 2

9 < 10

Hence,\(\frac{3}{5} < \frac{2}{3} \)

(d) \(\frac{3}{4} \square \frac{2}{8} \)

\(\frac{3}{4} and \frac{2}{8} \)

3 × 8 \(\square \) 4 × 2

24 < 8

Hence,\(\frac{3}{4} > \frac{2}{8} \)

(e) \(\frac{3}{5} \square \frac{5}{5} \)

\(\frac{3}{5} and \frac{5}{5} \)

3 × 5 \(\square \) 5 × 5

15 < 25

Hence,\(\frac{3}{5} < \frac{5}{5} \)

(f) \(\frac{7}{9} \square \frac{3}{9} \)

\(\frac{7}{9} and \frac{3}{9} \)

9 × 7 \(\square \) 9 × 3

63 > 27

Hence,\(\frac{7}{9} > \frac{3}{9} \)

(g) \(\frac{1}{4} \square \frac{2}{8} \)

\(\frac{1}{4} and \frac{2}{8} \)

1 × 8 \(\square \) 4 × 2

8 = 8

Hence,\(\frac{1}{4} = \frac{2}{8} \)

(h) \(\frac{6}{10} \square \frac{4}{5} \)

\(\frac{6}{10} and \frac{4}{5} \)

5 × 6 \(\square \) 10 × 4

30 < 40

Hence,\(\frac{6}{10} < \frac{4}{5} \)

(i) \(\frac{3}{4} \square \frac{7}{8} \)

\(\frac{3}{4} and \frac{7}{8} \)

3 × 8 \(\square \) 4 × 7

24 < 28

Hence,\(\frac{3}{4} < \frac{7}{8} \)

(j) \(\frac{6}{10} \square \frac{3}{5} \)

\(\frac{6}{10} and \frac{3}{5} \)

5 × 6 \(\square \) 10 × 3

30 = 30

Hence,\(\frac{6}{10} = \frac{3}{5} \)

(k) \(\frac{5}{7} \square \frac{15}{21} \)

\(\frac{5}{7} and \frac{15}{21} \)

5 × 21 \(\square \) 7 × 15

105 = 105

Hence,\(\frac{5}{7} = \frac{15}{21} \)

Q.6 The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.

(a) \(\frac{2}{12} \)

(b) \(\frac{3}{15} \)

(c) \(\frac{8}{50} \)

(d) \(\frac{16}{100} \)

(e) \(\frac{10}{60} \)

(f) \(\frac{15}{75} \)

(g) \(\frac{12}{60} \)

(h) \(\frac{16}{96} \)

(i) \(\frac{12}{75} \)

(j) \(\frac{12}{72} \)

(k) \(\frac{3}{18} \)

(l) \(\frac{4}{25} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{2}{12} = \frac{2÷2}{12÷2} = \frac{1}{6} \)

(b) \(\frac{3}{15} = \frac{3÷3}{15÷3} = \frac{1}{5} \)

(c) \(\frac{8}{50} = \frac{8÷2}{50÷2} = \frac{4}{25} \)

(d) \(\frac{16}{100} = \frac{16÷4}{100÷4} = \frac{4}{25} \)

(e) \(\frac{10}{60} = \frac{10÷10}{60÷10} = \frac{1}{6} \)

(f) \(\frac{15}{75} = \frac{15÷15}{75÷15} = \frac{1}{5} \)

(g) \(\frac{12}{60} = \frac{12÷12}{60÷12} = \frac{1}{5} \)

(h) \(\frac{16}{96} = \frac{16÷16}{196÷16} = \frac{1}{6} \)

(i) \(\frac{12}{75} = \frac{12÷3}{75÷3} = \frac{4}{25} \)

(j) \(\frac{12}{72} = \frac{12÷12}{72÷12} = \frac{1}{6} \)

(k)\(\frac{3}{18} = \frac{3÷3}{18÷3} = \frac{1}{6} \)

(l) \(\frac{4}{25} \)

Totally there are 3 groups of equivalent fractions.

\(\frac{1}{6} \)= (a), (e), (h), (j), (k)

\(\frac{1}{5} \) = (b), (f), (g)

\(\frac{4}{25} \) = (c), (d), (i), (l)

Q.7 Find answers to the following. Write and indicate how you solved them.

(a) Is \(\frac{5}{9} \) equal to \(\frac{4}{5} \)

(b) Is \(\frac{9}{16} \) equal to \(\frac{5}{9} \)

(c) Is \(\frac{4}{5} \) equal to \(\frac{16}{20} \)

(d) Is \(\frac{1}{15} \) equal to \(\frac{4}{30} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) Is \(\frac{5}{9} \) equal to \(\frac{4}{5} \)

by cross multiplication

5 × 5 \(\square \) 9 × 4

25 \(\neq \) 36

\(\frac{5}{9} \) \(\neq \) \(\frac{4}{5} \)

(b) Is \(\frac{9}{16} \) equal to \(\frac{5}{9} \)

by cross multiplication

9 × 9 \(\square \) 16 × 5

81 \(\neq \) 80

\(\frac{9}{16} \) \(\neq \) \(\frac{5}{9} \)

(c) Is \(\frac{4}{5} \) equal to \(\frac{16}{20} \)

by cross multiplication

20 × 4 \(\square \) 16 × 5

80 = 80

\(\frac{4}{5} \) \(\neq \) \(\frac{16}{20} \)

(d) Is \(\frac{1}{15} \) equal to \(\frac{4}{30} \)

by cross multiplication

30 × 1 \(\square \) 4 × 15

30 \(\neq \) 60

\(\frac{1}{15} \) \(\neq \) \(\frac{4}{30} \)

Q.8 Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac{2}{5} \) of the same book. Who read less?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Ila read 25 pages out of 100 pages

Fraction = \(\frac{25}{100} = \frac{25÷25}{100÷25} = \frac{1}{4} \)

Lalita read =\( \frac{2}{5} \)

\( \frac{2}{5} and \frac{1}{4} \)

By cross multiplication

4 × 2 and 5 ×1
8 > 5

\( \frac{2}{5} > \frac{1}{4} \)

Hence Lalita read less.

Q.9 Rafiq exercised for \(\frac{3}{6} \)of an hour, while Rohit exercised for \(\frac{3}{4} \) of an hour. Who exercised for a longer time?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Rafiq exercised for\(\frac{3}{6} \) of an hour.

Rohit exercised for \(\frac{3}{4} \) of an hour.

\(\frac{3}{6} \) and \(\frac{3}{4} \)

by cross multiplication

3 x 4 = 12 and 3 x 6 = 18

Since 12 < 18

\(\therefore \frac{3}{4} > \frac{3}{6} \)

Hence Rohit exercised for longer time.

Q.10 In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting with 60% or more marks?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

In class A, 20 students passed in first class out of 25 students.

\(\therefore \) Fraction of students getting first class

\(\frac{20}{25} = \frac{20÷5}{25÷5} = \frac{4}{5} \)

In class B, 24 students passed in first class out of 30 students.

\(\therefore \)Fraction of students getting first class

\(\frac{24}{30} = \frac{24÷6}{30÷6} = \frac{4}{5} \)

Comparing the two fractions, we get\(\frac{4}{5} = \frac{4}{5} \)

Hence, both the class A and B have the same fractions.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) The given figure represents the addition of

\(\frac{1}{5} + \frac{2}{5} = \frac{3}{5} \)

(b) The given figure represents the addition of

\(\frac{5}{5} + \frac{3}{5} = \frac{2}{5} \)

(c) The given figure represents the addition of

\(\frac{2}{6} + \frac{3}{6} = \frac{5}{6} \)

Q.2 Solve:

(a) \(\frac{1}{18} + \frac{1}{18} \)

(b) \(\frac{8}{15} + \frac{3}{15} \)

(c) \(\frac{7}{7} - \frac{5}{7} \)

(d) \(\frac{1}{22} + \frac{21}{22} \)

(e) \(\frac{12}{15} - \frac{7}{15} \)

(f) \(\frac{5}{8} + \frac{3}{8} \)

(g) 1 – \(\frac{2}{3} \) (1 = \(\frac{3}{3} \) )

(h) \(\frac{1}{4} + \frac{0}{4} \)

(i) 3 - \frac{12}{5} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{1}{18} + \frac{1}{18} \) = \(\frac{2}{18} = \frac{1}{9} \)

(b) \(\frac{8}{15} + \frac{3}{15} \) = \(\frac{11}{15} \)

(c) \(\frac{7}{7} - \frac{5}{7} \) = \(\frac{2}{7} \)

(d) \(\frac{1}{22} + \frac{21}{22} \) = \(\frac{22}{22} \) = 1

(e) \(\frac{12}{15} - \frac{7}{15} \) = \(\frac{5}{15} \) =
\(\frac{1}{5} \)

(f) \(\frac{5}{8} + \frac{3}{8} \) = \(\frac{8}{8} \) = 1

(g) 1 – \(\frac{2}{3} \) (1 = \(\frac{3}{3} \) )

= \(\frac{3}{3} \) - \(\frac{2}{3} \) = \(\frac{1}{3} \)

(h) \(\frac{1}{4} + \frac{0}{4} \) = \(\frac{1}{4} \)

(i) 3 - \frac{12}{5} \) = \(\frac{15-12}{5} \) = \(\frac{3}{5} \)

Q.3 Shubham painted \(\frac{2}{3} \) of the wall space in his room. His sister Madhavi helped and painted \(\frac{1}{3} \) of the wall space. How much did they paint together?

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

Fraction of wall painted by Shubham = \(\frac{2}{3} \)

Fraction of wall painted by Madhavi = \(\frac{1}{3} \)

Fraction of wall painted by Shubham and Madhavi

= \(\frac{2}{3} + \frac{1}{3} = \frac{3}{3} = 1 \)

\(\therefore \) Shubham and Madhavi together painted 1 complete wall in a room.

Q.4. Fill in the missing fractions.
(a) \(\frac{7}{10} - \square = \frac{3}{10} \)

(b)\(\square - \frac{3}{21} = \frac{5}{21} \)

(c) \(\square - \frac{3}{6} = \frac{3}{6} \)

(d) \(\square + \frac{5}{27} = \frac{12}{27} \)

NCERT Solutions for Class 6 Maths Chapter 7 Fractions

Answer :

(a) \(\frac{7}{10} \square = \frac{3}{10} \)

Or, \( - \square = \frac{3}{10} - \frac{7}{10} \)

Or, \(-\square = - \frac{4}{10} = -\frac{2}{5} \)

Or, \(\square = \frac{2}{5} \)

(b) \(\square - \frac{3}{21} = \frac{5}{21} \)

Or, \(\square = \frac{5}{21} + \frac{3}{21} \)

Or, \(\square = \frac{8}{21} \)

(c) \(\square - \frac{3}{6} = \frac{3}{6} \)

Or, \(\square = \frac{3}{6} + \frac{3}{6} \)

Or, \(\square = \frac{6}{6} = 1 \)
(d) \(\square + \frac{5}{27} = \frac{12}{27} \)

\(\square = \frac{12}{27} - \frac{5}{27} \)

\(\square = \frac{7}{27} \)

There are total 47 questions present in ncert solutions for class 6 maths chapter 7 fractions

There are total 4 long question/answers in ncert solutions for class 6 maths chapter 7 fractions

There are total 6 exercise present in ncert solutions for class 6 maths chapter 7 fractions